На движение по реке. катер один и тот же путь по течению реки за 2ч, а против течения за 2,45ч.скорость течения реки равна 2км/ч.найти собственную скорость катера!
1)Пусть x км/ч скорость катера. 2)Тогда x+2 км/ч скорость катера по течению,а x-2 км/ч-скорость катера против течения. 3)Следовательно,2(x+2)км-путь по течению,и 2,45(x-2)км-против течения. 4)Значит,2(x+2)=2,45(x-2) 2x+4=2,45x-4,9 2x+2,45x=(-4,9)-4 4,45x=(-8,9) x=(-8,9):(-4,45) x=2(км/ч)скорость катера. ответ:собственная скорость катера 2км/ч.
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тогда (х + 2) км/ч - скорость катера по течению реки,
(х - 2) км/ч - скорость катера против течения реки.
S = V * t
S₁ = S₂
V₁ * t₁ = V₂ * t₂
2(х + 2) = 2,45(х - 2)
2х + 4 = 2,45х - 4,9
2,45х - 2х = 4 + 4,9
0,45х = 8,9
х = 8,9 : 0,45
х = 19 7/9 (км/ч) - собственная скорость катера.
ответ: 19 7/9 км/ч.